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16t^2+18t+5=0
a = 16; b = 18; c = +5;
Δ = b2-4ac
Δ = 182-4·16·5
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{4}=2$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-2}{2*16}=\frac{-20}{32} =-5/8 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+2}{2*16}=\frac{-16}{32} =-1/2 $
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